Matemática, perguntado por TheJCL, 1 ano atrás

( IME ) Seja S = 1² + 3² + 5 ² + 7²...+79². O Valor de S = ?

Soluções para a tarefa

Respondido por Lukyo
6
\large\begin{array}{l} \textsf{S \'e a soma dos quadrados dos \'impares de 1 a 79:}\\\\ \mathsf{S=1^2+3^2+5^2+7^2+\ldots+79^2}\\\\ \mathsf{S=\displaystyle\sum_{k=1}^{40}(2k-1)^2}\\\\ \mathsf{S=\displaystyle\sum_{k=1}^{40}(4k^2-4k+1)\qquad(i)} \end{array}


\large\begin{array}{l} \textsf{Utilizando o conhecimento das somas telesc\'opicas, nossa}\\\textsf{inten\c{c}\~ao \'e reescrever o somando}\\\\ \mathsf{a_k=4k^2-4k+1}\\\\ \textsf{como combina\c{c}\~ao linear de polin\^omios, cuja soma \'e facilmente}\\\textsf{obtida pela propriedade telesc\'opica. Observe que}\\\\ \mathsf{a_k=4k^2-4k+1}\\\\ \mathsf{a_k=3\cdot (4k^2-4k+1)\cdot \dfrac{1}{3}}\\\\ \mathsf{a_k=\dfrac{1}{3}\cdot (12k^2-12k+3)}\\\\ \mathsf{a_k=\dfrac{1}{3}\cdot (4\cdot 3k^2-12k+3)}\\\\ \mathsf{a_k=\dfrac{1}{3}\cdot \left[4\cdot (3k^2+3k+1)-4(3k+1)-12k+3\right]}\\\\ \mathsf{a_k=\dfrac{1}{3}\cdot \left[4\cdot (3k^2+3k+1)-12k-4-12k+3\right]} \end{array}

\large\begin{array}{l} \mathsf{a_k=\dfrac{1}{3}\cdot \left[4\cdot (3k^2+3k+1)-24k-1\right]}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-8k-\dfrac{1}{3}}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-4\cdot \left(8k+\dfrac{1}{3}\right)\cdot \dfrac{1}{4}}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-4\cdot \left(2k+\dfrac{1}{12}\right)}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-4\cdot \left(2k+1-1+\dfrac{1}{12}\right)}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-4\cdot \left(2k+1-\dfrac{11}{12}\right)}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-4\cdot (2k+1)+4\cdot \dfrac{11}{12}} \end{array}

\large\begin{array}{l} \mathsf{a_k=\dfrac{4}{3}\cdot (3k^2+3k+1)-4\cdot (2k+1)+\dfrac{11}{3}}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot \left[(k+1)^3-k^3\right]-4\cdot \left[(k+1)^2-k^2\right]+\dfrac{11}{3}\cdot \left[(k+1)-k \right ]}\mathsf{\quad(ii)} \end{array}


\large\begin{array}{l} \textsf{Talvez voc\^e esteja a se perguntar o motivo de tantas manipula\c{c}\~oes}\\\textsf{simplesmente para reescrever}\\\\ \mathsf{a_k=4k^2-4k+1}\\\\ \textsf{como}\\\\ \mathsf{a_k=\dfrac{4}{3}\cdot \left[(k+1)^3-k^3 \right ]-4\cdot \left[(k+1)^2-k^2\right]+\dfrac{11}{3}\cdot \left[(k+1)-k \right ]}\\\\\\ \textsf{O motivo? Simples. Esta \'ultima forma \'e mais f\'acil de somar. :-)} \end{array}

_________


\large\begin{array}{l} \textsf{Vamos lembrar de uma propriedade bastate \'util dos somat\'orios:}\\\\ \mathsf{\displaystyle\sum_{k=p}^q (b_{k+1}-b_k)=-b_p+b_{q+1}}\\\\ \textsf{(propriedade telesc\'opica)}\\\\\\ \textsf{\'E f\'acil justificar por que isto funciona. Tente expandir os termos}\\\textsf{do somat\'orio e veja o que ocorre:}\\\\ \mathsf{\displaystyle\sum_{k=p}^q (b_{k+1}-b_k)}\\\\ =\mathsf{(b_{p+1}-b_p)+(b_{p+2}-b_{p+1})+\ldots+(b_{q+1}-b_q)}\\\\\\ \textsf{Haver\'a v\'arios cancelamentos de termos intermedi\'arios opostos,}\\\textsf{restando apenas}\\\\ =\mathsf{-b_p+b_{q+1}} \end{array}

_________


\large\begin{array}{l}\textsf{Voltando ent\~ao \`a nossa soma, temos que}\\\\ \mathsf{S=\displaystyle\sum_{k=1}^{40}(2k-1)^2}\\\\ =\mathsf{\displaystyle\sum_{k=1}^{40}(4k^2-4k+1)}\\\\ =\mathsf{\displaystyle\sum_{k=1}^{40}\,\left[\frac{4}{3}\left[(k+1)^3-k^3 \right ]-4\left[(k+1)^2-k^2\right]+\frac{11}{3}\big[(k+1)-k \big] \right ]}\\\\ =\mathsf{\displaystyle\frac{4}{3}\sum_{k=1}^{40}\left[(k+1)^3-k^3 \right ]-4\sum_{k=1}^{40}\left[(k+1)^2-k^2 \right ]+\frac{11}{3}\sum_{k=1}^{40}\big[(k+1)-k \big]} \end{array}

\large\begin{array}{l} =\mathsf{\displaystyle\frac{4}{3}\left[-1^3+(40+1)^3\right]-4\left[-1^2+(40+1)^2\right]+\frac{11}{3}\big[-1+(40+1)\big]}\\\\ =\mathsf{\displaystyle\frac{4}{3}\left[-1^3+41^3\right]-4\left[-1^2+41^2\right]+\frac{11}{3}\big[-1+41\big]}\\\\ =\mathsf{\displaystyle\frac{4}{3}\left[-1+68\,921\right]-4\left[-1+1\,681\right]+\frac{11}{3}\big[-1+41\big]}\\\\ =\mathsf{\displaystyle\frac{4}{3}\cdot 68\,920-4\cdot 1\,680+\frac{11}{3}\cdot 40} \end{array}

\large\begin{array}{l} =\mathsf{\dfrac{4}{3}\cdot 68\,920-\dfrac{12}{3}\cdot 1\,680+\dfrac{11}{3}\cdot 40}\\\\ =\mathsf{\dfrac{4\cdot 68\,920-12\cdot 1\,680+11\cdot 40}{3}}\\\\ =\mathsf{\dfrac{275\,680-20\,160+440}{3}}\\\\ =\mathsf{\dfrac{255\,520+440}{3}}\\\\ =\mathsf{\dfrac{255\,960}{3}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{S=85\,320} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}


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\large\begin{array}{l} \textsf{D\'uvidas comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}


Tags: soma sequência quadrados dos ímpares somatório telescópica cancelamentos simplificação cubo quadrado binômio de newton triângulo de pascal

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